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The one dimensional heat equation: Neumann and Robin ...

Neumann boundary ConditionsRobin boundary ConditionsThe one dimensional heat equation: Neumann and Robin boundary conditionsRyan C. DailedaTrinity UniversityPartial Differential EquationsFebruary 28, 2012 DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsThe heat equation with Neumann boundary conditionsOur goal is to solve:ut=c2uxx,0<x<L, 0<t,(1)ux(0,t) =ux(L,t) = 0,0<t,(2)u(x,0) =f(x),0<x<L.(3)As before, we will useseparation of variablesto find a family ofsimple solutions to (1) and (2), and then theprinciple ofsuperpositionto construct a solution satisfying (3).DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsSeparation of variablesAssuming thatu(x,t) =X(x)T(t), the heat equation (1) becomesXT =c2X impliesX X=T c2T=k,which we write asX kX= 0,(4)T c2kT= 0.(5)The initial conditions (2) becomeX (0)T(t) =X (L)T(t) = 0, orX (0) =X (L) = 0.(6)DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 1:k= 2>0 The ODE (4) becomesX 2X= 0 with general solutionX=c1e x+c2e boundary conditions (6) are0 =X (0) = c1 c2= (c1 c2),0 =X (L) = c1e L c2e L= (c1e L c2e L).

Neumann Boundary Conditions Robin Boundary Conditions Remarks At any given time, the average temperature in the bar is u(t) = 1 L Z L 0 u(x,t)dx. In the case of Neumann boundary conditions, one has u(t) = a 0 = f. That is, the average temperature is constant and is equal to the initial average temperature.

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Transcription of The one dimensional heat equation: Neumann and Robin ...

1 Neumann boundary ConditionsRobin boundary ConditionsThe one dimensional heat equation: Neumann and Robin boundary conditionsRyan C. DailedaTrinity UniversityPartial Differential EquationsFebruary 28, 2012 DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsThe heat equation with Neumann boundary conditionsOur goal is to solve:ut=c2uxx,0<x<L, 0<t,(1)ux(0,t) =ux(L,t) = 0,0<t,(2)u(x,0) =f(x),0<x<L.(3)As before, we will useseparation of variablesto find a family ofsimple solutions to (1) and (2), and then theprinciple ofsuperpositionto construct a solution satisfying (3).DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsSeparation of variablesAssuming thatu(x,t) =X(x)T(t), the heat equation (1) becomesXT =c2X impliesX X=T c2T=k,which we write asX kX= 0,(4)T c2kT= 0.(5)The initial conditions (2) becomeX (0)T(t) =X (L)T(t) = 0, orX (0) =X (L) = 0.(6)DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 1:k= 2>0 The ODE (4) becomesX 2X= 0 with general solutionX=c1e x+c2e boundary conditions (6) are0 =X (0) = c1 c2= (c1 c2),0 =X (L) = c1e L c2e L= (c1e L c2e L).

2 The first givesc1=c2. When we substitute this into the secondwe get2c1 sinh L= ,L>0, we must havec1=c2= 0. HenceX= 0, only trivial solutions in the case k> heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 2:k= 0 The ODE (4) is simplyX = 0 so thatX=Ax+ boundary conditions (6) yield0 =X (0) =X (L) = 1 we get the solutionX0= corresponding equation (5) forTisT = 0, which yieldsT=C. We setT0= give the zerothnormal mode:u0(x,t) =X0(x)T0(t) = heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 3:k= 2<0 The ODE (4) is nowX + 2X= 0 with solutionsX=c1cos x+c2sin boundary conditions (6) yield0 =X (0) = c1sin 0 + c2cos 0 = c2,0 =X (L) = c1sin L+ c2cos first of these givesc2= 0. In order forXto be nontrivial, thesecond shows that we also needsin L= heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 3:k= 2<0 This can occur if and only if L=n , that is = n=n L,n= 1, 2, 3, ..Choosingc1= 1 yields the solutionsXn= cos nx,n= 1,2,3.

3 For eachnthe corresponding equation (5) forTbecomesT = 2nT, with n=c n. Up to a constant multiple, thesolution isTn=e heat equationNeumann boundary ConditionsRobin boundary ConditionsNormal modes and superpositionMultiplying these together gives thenth normal modeun(x,t) =Xn(x)Tn(t) =e 2ntcos nx,n= 1,2,3, ..where n=n /Land n=c principle of superposition now guarantees that for any choiceof constantsa0,a1,a2, ..u(x,t) =a0u0+ n=1anun=a0+ n=1ane 2ntcos nx(7)is a solution of the heat equation (1) with the Neumann boundaryconditions (2).DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsInitial conditionsIf we now require that the solution (7) satisfy the initial condition(3) we find that we needf(x) =u(x,0) =a0+ n=1ancosn xL,0<x< is simply thecosine series expansion of f(x). Using ourprevious results, we finally find that iff(x) is piecewise smooththena0=1L L0f(x)dx,an=2L L0f(x) cosn xLdx,n heat equationNeumann boundary ConditionsRobin boundary ConditionsConclusionTheoremIf f(x)is piecewise smooth, the solution to the heat equation(1)with Neumann boundary conditions (2)and initial conditions (3)isgiven byu(x,t) =a0+ n=1ane 2ntcos nx,where n=n L, n=c n,and the coefficients a0,a1,a2.

4 Are those occurring in the cosineseries expansion of f(x). They are given explicitly bya0=1L L0f(x)dx,an=2L L0f(x) cosn xLdx,n heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 1 ExampleSolve the following heat conduction problem:ut=14uxx,0<x<1,0<t,ux(0,t) =ux(1,t) = 0,0<t,u(x,0) = 100x(1 x),0<x< 1, n=n andf(x) = 100x(1 x) we finda0= 10100x(1 x)dx=503an= 2 10100x(1 x) cosn x dx= 200(1 + ( 1)n)n2 2,n heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 1 Sincec= 1/2, n=n /2. Plugging everything into our generalsolution we getu(x,t) =503 200 2 n=1(1 + ( 1)n)n2e n2 2t/4cosn in the case of Dirichlet boundary conditions , the exponentialterms decay rapidly witht. We havelimt u(x,t) = heat equationNeumann boundary ConditionsRobin boundary ConditionsRemarksAt any given time, theaverage temperaturein the bar isu(t) =1L L0u(x,t) the case of Neumann boundary conditions , one hasu(t) =a0= is,the average temperature is constantandis equal tothe initial average in this caselimt u(x,t) =a0for allx.

5 That is, at any point in the barthe temperaturetends to the initial average heat equationNeumann boundary ConditionsRobin boundary ConditionsThe heat equation with Robin boundary conditionsWe now consider the problemut=c2uxx,0<x<L, 0<t,u(0,t) = 0,0<t,(8)ux(L,t) = u(L,t),0<t,(9)u(x,0) =f(x),0<x< (9) we take >0. This states that the bar radiates heat toits surroundings at a rate proportional to its that conditions such as (9) are heat equationNeumann boundary ConditionsRobin boundary ConditionsSeparation of variablesAs before, the assumption thatu(x,t) =X(x)T(t) leads to theODEsX kX= 0,T c2kT= 0,and the boundary conditions (8) and (9) implyX(0) = 0,X (L) = X(L).Also as before, the possibilities forXdepend on the sign of theseparation heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 1:k= 0We haveX = 0 and soX=Ax+Bwith0 =X(0) =B,A=X (L) = X(L) = (AL+B).Together these giveA(1 + L) = 0. Since ,L>0, we haveA= 0and henceX= 0.

6 Thus,there are only trivial solutions in this heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 2:k= 2>0 Once again we haveX 2X= 0 andX=c1e x+c2e boundary conditions become0 =c1+c2, (c1e L c2e L) = (c1e L+c2e L).The first givesc2= c1, which when substituted in the secondyields2 c1cosh L= 2 c1sinh heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 2:k= 2>0We may rewrite this asc1( cosh L+ sinh L) = quantity in parentheses is positive (since , andLare), sothis means we must havec1= c2= 0. HenceX= 0 andthere are only trivial solutions in this heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 3:k= 2<0 FromX + 2X= 0 we findX=c1cos x+c2sin xand from the boundary conditions we have0 =c1, ( c1sin L+c2cos L) = (c1cos L+c2sin L).Together these imply thatc2( cos L+ sin L) = heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 3:k= 2<0 Since we want nontrivial solutions ( 0), we must have cos L+ sin L= 0which can be rewritten astan L=.

7 This equation has an infinite sequence of positive solutions0< 1< 2< 3< DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsCase 3:k= 2<0 The figure below shows the curvesy= tan L(in red) andy= / (in blue).The -coordinates of their intersections (in pink) are the values 1, 2, 3, ..DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsRemarksFrom the diagram we see that:For eachn(2n 1) /2L< n<n n (2n 1) values of andLtend to accelerate this heat equationNeumann boundary ConditionsRobin boundary ConditionsNormal modesAs in the earlier situations, for eachn 1 we have the solutionXn= sin nxand the correspondingTn=e 2nt, n=c nwhich give thenormal modesof the heat equation with boundaryconditions (8) and (9)un(x,t) =Xn(x)Tn(t) =e 2ntsin heat equationNeumann boundary ConditionsRobin boundary ConditionsSuperpositionSuperposition of normal modes gives thegeneral solutionu(x,t) = n=1cnun(x,t) = n=1cne 2ntsin the initial conditionu(x,0) =f(x) gives usf(x) = n=1cnsin is ageneralized Fourier seriesforf(x).

8 It isdifferentfromthe ordinary sine series forf(x) since nis not a multiple of a common heat equationNeumann boundary ConditionsRobin boundary ConditionsGeneralized Fourier coefficientsTo compute thegeneralized Fourier coefficientscnwe will usethe following functionsX1(x) = sin 1x,X2(x) = sin 2x,X3(x) = sin 3x, ..form a complete orthogonal set on[0,L].Completemeans that all sufficiently nice functions can berepresented via generalized Fourier series. This is aconsequence ofSturm-Liouville theory, which we will can verify orthogonality directly, and will use this toexpress the coefficientscnas ratios of inner heat equationNeumann boundary ConditionsRobin boundary ConditionsGeneralized Fourier coefficientsAssuming orthogonality for the moment, for anyn 1 we have thefamiliar computationhf,Xni= m=1cmsin mx,sin nx = m=1cmhsin mx,sin nxi=cnhsin nx,sin nxi=cnhXn,Xnisince the inner products withm6=nall equal heat equationNeumann boundary ConditionsRobin boundary ConditionsGeneralized Fourier coefficientsIt follows immediately that the generalized Fourier coefficients aregiven bycn=hf,XnihXn,Xni= L0f(x) sin nx dx L0sin2 nx any givenf(x) these integrals can typically be computedexplicitly in terms of values of n, however, must typically be found vianumerical heat equationNeumann boundary ConditionsRobin boundary ConditionsConclusionTheoremThe solution to the heat equation(1)

9 With Robin boundaryconditions(8)and(9)and initial condition(3)is given byu(x,t) = n=1cne 2ntsin nx,where nis the nth positive solution totan L= , n=c n, and the coefficients cnare given bycn= L0f(x) sin nx dx L0sin2 nx heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 2 ExampleSolve the following heat conduction problem:ut=125uxx,0<x<3,0<t,u(0,t) = 0,0<t,ux(3,t) = 12u(3,t),0<t,u(x,0) = 100(1 x3),0<x< havec= 1/5,L= 3, = 1/2 andf(x) = 100(1 x/3).DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 2 The integrals defining the Fourier coefficients are100 30(1 x3)sin nx dx=100(3 n sin 3 n)3 2nand 30sin2 nx dx=32+ cos23 (3 n sin 3 n)3 2n(3 + 2 cos23 n).DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 2We can therefore write out the full solution asu(x,t) = n=1200(3 n sin 3 n)3 2n(3 + 2 cos23 n)e 2nt/25sin nx,where nis thenth positive solution to tan 3 = 2 .DailedaThe heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 2 Remarks:In order to use this solution for numerical approximation orvisualization, we must compute the values can be done numerically in Maple, using thefsolvecommand.

10 Specifically, ncan be computed via the commandfsolve(tan(m L)=-m/k,m=(2 n-1) Pi/(2 L)..n Pi/L);whereLandkhave been assigned the values ofLand , values can be computed and stored in anArraystructure, or one can define nas a function using the-> heat equationNeumann boundary ConditionsRobin boundary ConditionsExample 2 Here are approximate values for the first 5 values of (x,t) = ( ) + ( )+ ( ) + ( )+ ( ) + DailedaThe heat equatio


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